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                <div class="bbp-reply-header" id="post-218117">
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                  <span class="bbp-reply-post-date">
                   2007年1月17日 下午1:40
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3662/">
                    为什么kernel estimate能用到fft呢？
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                  <a class="bbp-reply-permalink" href="http://cos.name/cn/topic/3662/#post-218117">
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                   <p>
                    [b]引用第2楼[i]rtist[/i]于[i]2007-01-17 20:01[/i]发表的“”[/b]:
                    <br/>
                    [url]http://www.jstor.org/view/00359254/di993351/99p0018j/0[/url]
                   </p>
                  </blockquote>
                  <p>
                   里面一个关键的东西看不懂。
                  </p>
                  <p>
                   <img src="http://maths.utime.cn:81/textool/cgi-bin/nictex.exe?u(s)=(2\pi)^{-1/2}n^{-1}\sum{e^{isXj}}"/>
                  </p>
                  <p>
                   为什么上面这个式子是原始数据的FT呢？原始数据是离散的，做的应该是DFT啊，但是式子写出来又不像DFT。紧接着文章说
                  </p>
                  <blockquote class="d4pbbc-quote">
                   <p>
                    A discrete approximation to u(s) is found by constructing a histogram on a grid of 2^k cells and then applying the Fast Fourier transform;
                   </p>
                  </blockquote>
                  <p>
                   这些cells是怎么构造的？这里histogram是指原始数据落在cells的频数吗？这样就能计算出u(s)的近似值了？
                  </p>
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                <div class="bbp-reply-header" id="post-218113">
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                   2007年1月17日 下午12:05
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3662/">
                    为什么kernel estimate能用到fft呢？
                   </a>
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                   <p>
                    [b]引用第2楼[i]rtist[/i]于[i]2007-01-17 20:01[/i]发表的“”[/b]:
                    <br/>
                    [url]http://www.jstor.org/view/00359254/di993351/99p0018j/0[/url]
                   </p>
                  </blockquote>
                  <p>
                   哎，我应该想到先去期刊上找的。我去研究一下。谢谢！
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                <div class="bbp-reply-header" id="post-218112">
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                   2007年1月17日 下午12:03
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3662/">
                    为什么kernel estimate能用到fft呢？
                   </a>
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                  <a class="bbp-reply-permalink" href="http://cos.name/cn/topic/3662/#post-218112">
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                    [b]引用第1楼[i]rtist[/i]于[i]2007-01-17 19:58[/i]发表的“”[/b]:
                    <br/>
                    选择窗宽?
                   </p>
                  </blockquote>
                  <p>
                   不像。我贴一下KernSmooth里面叫做bkde的程序
                  </p>
                  <pre class="highlight ">function (x, kernel = "normal", canonical = FALSE, bandwidth, 
    gridsize = 401, range.x, truncate = TRUE) 
{
    n &lt;- length(x)
    M &lt;- gridsize
    if (kernel == "normal") 
        del0 &lt;- (1/(4 * pi))^(1/10)
    if (kernel == "box") 
        del0 &lt;- (9/2)^(1/5)
    if (kernel == "epanech") 
        del0 &lt;- 15^(1/5)
    if (kernel == "biweight") 
        del0 &lt;- 35^(1/5)
    if (kernel == "triweight") 
        del0 &lt;- (9450/143)^(1/5)
    if (missing(bandwidth)) {
        if (canonical) {
            bandwidth &lt;- (243/(35 * n))^(1/5) * sqrt(var(x))
        }
        else {
            bandwidth &lt;- del0 * (243/(35 * n))^(1/5) * sqrt(var(x))
        }
    }
    h &lt;- bandwidth
    if (canonical) {
        if (kernel == "normal") {
            tau &lt;- 4 * del0
        }
        else {
            tau &lt;- del0
        }
    }
    else {
        if (kernel == "normal") {
            tau &lt;- 4
        }
        else {
            tau &lt;- 1
        }
    }
    if (missing(range.x)) {
        range.x &lt;- c(min(x) - tau * h, max(x) + tau * h)
    }
    a &lt;- range.x[1]
    b &lt;- range.x[2]
    gpoints &lt;- seq(a, b, length = M)
    gcounts &lt;- linbin(x, gpoints, truncate)
    L &lt;- min(floor(tau * h * (M - 1)/(b - a)), M)
    lvec &lt;- (0:L)
    delta &lt;- (b - a)/(h * (M - 1))
    if (canonical == FALSE) 
        del0 &lt;- 1
    if (kernel == "normal") {
        kappa &lt;- dnorm(lvec * delta/del0)/(n * h * del0)
    }
    else if (kernel == "box") {
        kappa &lt;- 0.5 * dbeta(0.5 * (lvec * delta/del0 + 1), 1, 
            1)/(n * h * del0)
    }
    else if (kernel == "epanech") {
        kappa &lt;- 0.5 * dbeta(0.5 * (lvec * delta/del0 + 1), 2, 
            2)/(n * h * del0)
    }
    else if (kernel == "biweight") {
        kappa &lt;- 0.5 * dbeta(0.5 * (lvec * delta/del0 + 1), 3, 
            3)/(n * h * del0)
    }
    else if (kernel == "triweight") {
        kappa &lt;- 0.5 * dbeta(0.5 * (lvec * delta/del0 + 1), 4, 
            4)/(n * h * del0)
    }
    P &lt;- 2^(ceiling(log(M + L)/log(2)))
    kappa &lt;- c(kappa, rep(0, P - 2 * L - 1), kappa[(L + 1):2])
    gcounts &lt;- c(gcounts, rep(0, P - M))
    kappa &lt;- fft(kappa)
    gcounts &lt;- fft(gcounts)
    return(list(x = gpoints, y = (Re(fft(kappa * gcounts, TRUE))/P)[1:M]))
}
&lt;environment: namespace:KernSmooth&gt;

</pre>
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                <div class="bbp-reply-header" id="post-218108">
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                  <span class="bbp-reply-post-date">
                   2007年1月17日 上午11:15
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3658/">
                    《Mathematical Statistics and Data Analysis》此书有没有中文版？
                   </a>
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                   <p>
                    [b]引用第3楼[i]jackson[/i]于[i]2007-01-17 12:24[/i]发表的“”[/b]:
                    <br/>
                    你看陈家鼎的《数理统计学讲义》吧，有新出的第二版，很不错的呀
                   </p>
                  </blockquote>
                  <p>
                   我刚上过这门课。课本就用这个，但是没觉得有多好，可能是风格不太适合我。上这课的时候不懂的地方基本都去参考《Mathematical Statistics and Data Analysis》，还是这本比较经典。
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                <div class="bbp-reply-header" id="post-217581">
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                   2007年1月9日 上午4:11
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3366/">
                    如何用随机模拟验证均匀分布下，均值和标准差不独立？
                   </a>
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                  <a class="bbp-reply-permalink" href="http://cos.name/cn/topic/3366/#post-217581">
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                   <p>
                    [b]引用第5楼[i]rtist[/i]于[i]2007-01-09 03:04[/i]发表的“”[/b]:
                    <br/>
                    越看越糊涂，不都已经证出来了么？到底还要验证什么？没什么好模拟的啊？
                   </p>
                  </blockquote>
                  <p>
                   就是想再用数据验证一下。均匀分布的时候做出来的均值和标准差的散点图看上去确实比较像独立的。
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                  <span class="bbp-reply-post-date">
                   2007年1月8日 下午1:48
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3403/">
                    求助:响应面分析的问题(解决者给予酬劳)
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                  <blockquote class="d4pbbc-quote">
                   <p>
                    [b]引用第2楼[i]谢益辉[/i]于[i]2007-01-08 16:04[/i]发表的“”[/b]:
                    <br/>
                    里面怎么有那么多-1？……
                   </p>
                   <p>
                    做了个回归，发现效果不好哦：）
                   </p>
                  </blockquote>
                  <p>
                   我也做了，结论是所有变量基本上都找不到"响应"。用交叉验证线性回归，结果如下
                  </p>
                  <pre class="highlight "> Validation selection criteria:
      (Intercept) dat1[, -5]1 dat1[, -5]2 dat1[, -5]3 dat1[, -5]4       cv
 [1,]           1           0           0           0           0 0.002421
 [2,]           1           0           1           0           0 0.002591
 [3,]           1           1           0           0           0 0.002699
 [4,]           1           0           0           0           1 0.002805
 [5,]           1           0           0           1           0 0.002902
 [6,]           1           1           1           0           0 0.002932
 [7,]           1           0           1           0           1 0.003019
 [8,]           1           0           1           1           0 0.003093
 [9,]           1           1           0           0           1 0.003143
[10,]           1           1           0           1           0 0.003194
[11,]           1           0           0           1           1 0.003345
[12,]           1           1           1           0           1 0.003439
[13,]           1           1           1           1           0 0.003458
[14,]           1           0           1           1           1 0.003600
[15,]           1           1           0           1           1 0.003700
[16,]           1           1           1           1           1 0.004059
[17,]           0           0           0           0           1 0.024530
[18,]           0           1           0           0           0 0.024910
[19,]           0           0           0           1           0 0.025540
[20,]           0           0           1           0           0 0.025570

Printed the first  20  best models 
</pre>
                  <p>
                   仅对因素2做线性回归，结果如下
                  </p>
                  <pre class="highlight ">Residuals:
      Min        1Q    Median        3Q       Max 
-0.094667 -0.039333  0.001333  0.037833  0.063333 

Coefficients:
             Estimate Std. Error t value Pr(&gt;|t|)    
(Intercept)  0.141667   0.009029  15.690 1.89e-14 ***
dat1[, 2]   -0.015667   0.013544  -1.157    0.258    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 0.04692 on 25 degrees of freedom
Multiple R-Squared: 0.0508,     Adjusted R-squared: 0.01284 
F-statistic: 1.338 on 1 and 25 DF,  p-value: 0.2583 
</pre>
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                   2007年1月8日 下午1:39
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3366/">
                    如何用随机模拟验证均匀分布下，均值和标准差不独立？
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                   <p>
                    [b]引用第3楼[i]pigtail[/i]于[i]2007-01-08 18:34[/i]发表的“”[/b]:
                   </p>
                   <p>
                    连续的从均值和标准差的表达式就可以看出不独立，若要随机模拟，可以检验他们相关，从而得出不独立。
                   </p>
                  </blockquote>
                  <p>
                   正态的时候独立。
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                <div class="bbp-reply-header" id="post-217467">
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                   2007年1月8日 上午2:45
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3366/">
                    如何用随机模拟验证均匀分布下，均值和标准差不独立？
                   </a>
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                   <p>
                    [b]引用第1楼[i]rtist[/i]于[i]2007-01-08 02:05[/i]发表的“”[/b]:
                    <br/>
                    Unif(a,b)
                   </p>
                   <p>
                    mean=(a+b)/2
                    <br/>
                    var=(b-a)^2/12=(mean-a)^2/3
                   </p>
                  </blockquote>
                  <p>
                   独立性检验。
                  </p>
                  <p>
                   离散时可以用卡方检验，连续呢？
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                   2007年1月3日 下午3:21
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3296/">
                    E(X|Y)=E(X)是否可以推出X,Y相互独立？
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                  <p>
                   谢谢！
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                   2007年1月1日 下午3:39
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                   回复：
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                    如何让一个程序重复运行500次？（谢谢）
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                   <p>
                    [b]引用第7楼[i]rtist[/i]于[i]2006-12-30 17:30[/i]发表的“”[/b]:
                   </p>
                   <p>
                    其实这东西根本没啥用，就是给我这种懒人预备的，运行速度上通常比不上显示循环。
                    <br/>
                    我常用的也就是apply,sapply,lapply,和papply了；有人也用tapply，我几乎从来不用。
                   </p>
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                  <p>
                   真的么？我还以为要比循环快，在拼命的用啊。
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                   2006年12月20日 下午2:39
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3074/">
                    我写的R调用DLL的最简单例子，也许对大家有点用处
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                  <p>
                   考完试一定细看。
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                <div class="bbp-reply-header" id="post-215963">
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                   2006年12月20日 上午8:05
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                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3068/">
                    要期末考试了，平时偷懒的人赶快学习……
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                  <p>
                   刚赶完两份数值模拟的大作业。还有n份试验报告要赶。
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                <div class="bbp-reply-header" id="post-215711">
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                   2006年12月17日 上午8:00
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/3004/">
                    求助：如何处理下述的字符读取问题
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                   是啊。强！
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                   2006年12月16日 上午7:35
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                   回复：
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                    求助：如何处理下述的字符读取问题
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                   drewlee
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                   普通会员
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                   <p>
                    [b]引用第1楼[i]谢益辉[/i]于[i]2006-12-16 14:42[/i]发表的“”[/b]:
                    <br/>
                    先用read.table直接读进来，为了防止最后一行的长度不一样，可以加上fill=TRUE参数，然后as.matrix，转化为字符矩阵，再用sub/gsub之类的函数替换掉*:，这里的正则表达式（regular expression）怎么写我还没搞明白，最后字符型的数字就提取出来了，用as.numeric可以转化为数值型数据。
                   </p>
                  </blockquote>
                  <p>
                   Thanks!通过你的提示我找到了解决方案。
                  </p>
                  <p>
                   搞定了。可能不够简洁。如下：
                  </p>
                  <pre class="highlight ">xx=scan("file.txt",what="character")
xxx=as.numeric(as.vector(as.matrix(data.frame(strsplit(xx,":"))[2,])))
iii=as.numeric(as.vector(as.matrix(data.frame(strsplit(xx,":"))[1,])))

raw.p=xxx[order(iii)]
</pre>
                  <p>
                   上面是按照冒号前面的数字排的，和开始我说得不太一样。
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                <div class="bbp-reply-header" id="post-215524">
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                  <span class="bbp-reply-post-date">
                   2006年12月15日 上午5:00
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                   回复：
                   <a class="bbp-topic-permalink" href="http://cos.name/cn/topic/2991/">
                    想把一个数值的矩阵映射为一个颜色方格的矩阵，什么函数？
                   </a>
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                  <a class="bbp-reply-permalink" href="http://cos.name/cn/topic/2991/#post-215524">
                   3 楼
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                   <p>
                    [b]引用第1楼[i]谢益辉[/i]于[i]2006-12-15 12:02[/i]发表的“”[/b]:
                    <br/>
                    image(x=1:5,y=1:5,z=matrix(25:1,5),col=heat.colors(25))
                   </p>
                   <p>
                    只是一个示例，具体参数你自己调整吧：）
                   </p>
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                  <p>
                   Thanks!
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